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I read this MSE answer on why the unit group of $\mathbb{Z}/p^n\mathbb{Z}$ for odd prime $p$ is cyclic. I do understand that $\exp_p$ is an isomorphism from $p\mathbb{Z}_p$ (additive) to $1 + p\mathbb{Z}_p$ (multiplicative). However, I do not understand the exact steps paul garret did (how do they conclude that the quotient must be cyclic? what does index-$p$ part mean? what is a (pro-) generator?). I would greatly appreciate it if someone could provide a more elaborate and detailed answer, or provide some more (different) $p$-adic proofs.

Jonas Hardt
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  • I do not think he uses the $p$-adic exponential to show that the unit group of $\mathbb{Z}/p\mathbb{Z}$ is cyclic. It is rather the opposite way. – Maty Mangoo Aug 11 '23 at 21:18
  • A $p$-adic proof not using the $p$-adic exponential is Theorem 1.1 of https://kconrad.math.uconn.edu/blurbs/gradnumthy/primepowerunitsandGLnQ.pdf. The proof and a nice corollary are in Section 2. The key $p$-adic property is $|a^n - 1|_p = |n|_p|a-1|_p$ when $p \not= 2$ and $|a-1|_p < 1$. Remark 2.2 there presents a more conceptual argument with the $p$-adic logarithm, which is the kind of argument you are asking about. – KCd Aug 11 '23 at 21:18
  • Cf. https://math.stackexchange.com/a/3221919/96384 for a proof that hides its $p$-adic underpinnings. – Torsten Schoeneberg Aug 11 '23 at 21:19
  • One can check quite elementary that the unit group of $\mathbb{Z}/p^n\mathbb{Z}$ is cyclic; it decomposes as $$(\mathbb{Z}/p^n\mathbb{Z})^{\times} \cong \mathbb{Z}/p^{n-1}\mathbb{Z} \times \mathbb{Z}/(p-1)\mathbb{Z},$$ and by the CRT (chin.rem.th.), this is cyclic. – Maty Mangoo Aug 11 '23 at 21:20
  • @MatyMangoo a typo in your previous comment: the modulus on the left side of the displayed isomorphism should be $p^n$, not $p$. – KCd Aug 11 '23 at 21:21
  • @KCd that was quick! :) Thanks. – Maty Mangoo Aug 11 '23 at 21:22
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    @MatyMangoo I disagree that cyclicity of the units mod $p^n$ when $p > 2$ is "quite elementary" to someone seeing that for the first time. I agree that once you know there's such a decomposition then the unit group mod $p^n$ is cyclic but what is a source of that direct product decomposition? You could show $1+p \bmod p^n$ has order $p^{n-1}$, but showing there's a cyclic subgroup with order $p-1$ is not so easy to see at an elementary level: it needs some careful details to check. – KCd Aug 11 '23 at 21:24

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